Lee–Kesler method

${\displaystyle \ln P_{r}=f^{(0)}+\omega \cdot f^{(1)}}$

${\displaystyle f^{(0)}=5.92714-{\frac {6.09648}{T_{r}}}-1.28862\cdot \ln T_{r}+0.169347\cdot T_{r}^{6}}$

${\displaystyle f^{(1)}=15.2518-{\frac {15.6875}{T_{r}}}-13.4721\cdot \ln T_{r}+0.43577\cdot T_{r}^{6}}$

with

${\displaystyle P_{r}={\frac {P}{P_{c}}}}$ (reduced pressure) and ${\displaystyle T_{r}={\frac {T}{T_{c}}}}$ (reduced temperature).

The prediction error can be up to 10% for polar components and small pressures and the calculated pressure is typically too low. For pressures above 1 bar, that means, above the normal boiling point, the typical errors are below 2%. [2]

For benzene with

• T_c = 562.12 K[3]
• P_c = 4898 kPa[3]
• T_b = 353.15 K[4]
• ω = 0.2120[5]

the following calculation for T=T_b results:

• T_r = 353.15 / 562.12 = 0.628247
• f⁽⁰⁾ = -3.167428
• f⁽¹⁾ = -3.429560
• P_r = exp( f⁽⁰⁾ + ω f⁽¹⁾ ) = 0.020354
• P = P_r * P_c = 99.69 kPa

The correct result would be P = 101.325 kPa, the normal (atmospheric) pressure. The deviation is -1.63 kPa or -1.61 %.

It is important to use the same absolute units for T and T_c as well as for P and P_c. The unit system used (K or R for T) is irrelevant because of the usage of the reduced values T_r and P_r.

• Vapour pressure of water
• Antoine equation
• Tetens equation
• Arden Buck equation
• Goff–Gratch equation